Equation Of A Line Concise Solutions Chapter-14 Class 10

Selina Publications Chapter-14 Equation of a Straight Line

Equation Of A Line Concise Solutions Chapter-14 Class 10. Solutions of Exercise – 14 (A), Exercise – 14 (B),Exercise – 14 (C), Exercise – 14 (D),Exercise – 14 (E) of Concise.  Concise Selina Maths of ICSE Board Class 10th Solutions for Equation Of A Line Concise Solutions Chapter-14. Step by Step Solutions of Concise Equation Of A LineChapter-14 for ICSE Maths Class 10 is available here. All Solutions of Concise Selina of Chapter-14 Equation Of A Line has been solved according instruction given by council. This is the  Solutions of Chapter-14 Equation Of A Line for ICSE Class 10th. ICSE Maths text book of Concise is In series of famous ICSE writer in maths publications. Concise is most famous among students.

Equation Of A Line Concise Solutions Chapter-14 Class 10

The Solutions of Concise Mathematics Equation Of A Line Concise Solutions Chapter-14 for ICSE Class 10 have been solved.  Experience teachers Solved Equation Of A Line Concise Solutions Chapter-14 to help students of class 10th ICSE board. Therefore the ICSE Class 10th Maths Solutions of Concise Selina Publishers helpful on  various topics which are prescribed in most ICSE Maths textbook

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Exe-14(A),    Exe-14(B),    Exe-14(C)   Exe-14(D),    Exe-14(E)

How to Solve Concise Maths Selina Publications Chapter-14 Equation Of A Line

Note:- Before viewing Solutions of Chapter -14 Equation Of A Line of Concise Selina Maths. Read the Chapter Carefully then solve all example of your text bookThe Chapter- 14 Equation Of A Line is main Chapter in ICSE board.

Concise Solutions of Exercise -14(A),Equation of a line for ICSE Class 10th Maths 

Question 1.

Find, which of the following points lie on the line x – 2y + 5 = 0 :
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)

Answer 1

Equation of given line x – 2y + 5 = 0
(i)

Substituting x = 1, y = 3, in the given equation.
1 – 2 x 3 + 5 = 0 ⇒ 1 – 6 + 5 = 0 ⇒ 0 = 0, which is true.
(1, 3) satisfies the equation.
(ii)

Substituting x = 0 , y = 5 in the given equation
0 – 2 x 5 + 5 = 0 ⇒ 0 – 10 + 5 = 0 ⇒ -5 = 0, which is not true.
( 0, 5) does not satisfy the equation.

(iii)

 

Substituting x = – 5, y = 0 in the given equation
-5 – 2 x 0 + 5 = 0 ⇒ -5 – 0 + 5 = 0 ⇒ 0 = 0 which is true.
(-5, 0) satisfies the equation.

(iv)

Substituting x = 5, y = 5 in the given equation.
– 5 – 2 x 5 + 5 = 0 ⇒ -5 – 10 + 5 = 0 ⇒ 0 = 0 which is true.
(5, 5) satisfies the equation.

(v)

Substituting x = 2, y = -1.5 in the given equation.
2 – 2 x (- 1.5) + 5 = 0 ⇒ 2 + 3 + 5 = 0 ⇒ 10 = 0. which is not true.
(2, -1.5) does not satisfy the equation.

(vi)

Substituting x = -2, y = -1.5 in the given equation
– 2 – 2 x (-1.5) + 5 = 0 ⇒ – 2 + 3 + 5 = 0 ⇒ 6 = 0, which is not true.
(-2, -1.5) does not satisfies the equation.

Question 2

State, true or false :
(i) the line \frac { x }{ 2 } + \frac { y }{ 3 } = 0 passes through the point (2, 3).
(ii) the line \frac { x }{ 2 } + \frac { y }{ 3 } = 0 passes through the point (4, -6).
(iii) the point (8, 7) lies on the line y – 7 = 0
(iv) the point (-3, 0) lies on the line x + 3 = 0
(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.
Solution:

Answer 2

(i)

Equation of the line is \frac { x }{ 2 } + \frac { y }{ 3 } = 0
and co-ordinates of point are (2, 3)
If the point is on the line, then it will satisfy the equation.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 2
(2, 3) is not on the line

(ii)

Equation of the line is \frac { x }{ 2 } + \frac { y }{ 3 } = 0
and co-ordinates of point are (4, -6)
If the point is on the line, then it will satisfy the equation
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 2.1
Hence, point (4, -6) is on the line.

(iii)

Equation of line is y – 7 = 0 and the co-ordinates of point are (8, 7)
If the point is on the line, then it will satisfy the equation
L.H.S. = y – 7 = 7 – 7 = 0 = R.H.S.
Hence, point (8, 7) is on the line.

(iv)

Equation of the line is x + 3 = 0 and co-ordinates of point are (-3, 0)
If the point is on the line, then it will satisfy the equation.
L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S.
Hence, the point (-3, 0) is on the line.

(v)

Equation of the line is 2x – y = 3
and co-ordinates of the point are (2, a)
If the point is on the line, then it will satisfy the equation.
L.H.S. = 2x – y = 2 x 2 – a = 4 – a
R.H.S. = 3
4 – a = 3 ⇒ 4 + 3 = a ⇒ a = 7
But a = 5 given, therefore it is not on the line.
(i) False

(ii) True

(iii) True

(iv) True

(v) False.

Question 3

The line given by the equation 2x – \frac { y }{ 3 } = 7 passes through the point (k, 6); calculate the value of k.

Answer 3

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 3

Question 4

For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?

Answer  4

Point (3, -k) satisfies the equation 9x + 4y = 3
Substituting x = 3 , y = -k, we get :
9 x 3 + 4 (- k), = 3
⇒ 27 – 4k = 3
and ⇒ – 4k = 3 – 27
so ⇒ – 4k = – 24
hence ⇒ k = 6

Question 5

The line \frac { 3x }{ 5 } – \frac { 2y }{ 3 } + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.

Answer 5

Point (m, 2m -1) satisfies the equation
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 5.1

Question 6

Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?

Answer 6

Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2)
The mid-point of (5, -2) and (-1, 2) satisfies the equation.
Now, mid-point of (5, -2) and (-1, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 6
Now, substituting x = 2, y = 0, in the given equation

Question 7

(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of k.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and ( 0, k). Find the value of k.

Answer 7

(i) line y = 3x – 2 bisects the join of (a, 3) and (2, -5)
Mid-point join of there points satisfies it.
Now, mid-point of (a, 3) and (2, -5) is
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 7.1

Question 8

(i) The point (-3, -2) lies on the line ax + 3y + 6 = 0, calculate the value of ‘a’
(ii) The line y = mx + 8 contains the point (- 4, 4), calculate the value of ‘m’

Answer  8

(i) Point (-3, 2) lies on the line ax + 3y + 6 = 0,
Then x = – 3, y = 2 satisfies it
a (-3) + 3(2) + 6 = 0
⇒ -3a + 6 + 6 = 0
so ⇒ -3a + 12 = 0
therefore ⇒ -3a = – 12
hence ⇒ a = 4
(ii) line y = mx + 8 contains the point (-4, 4)
x = – 4, y = 4 satisfies it
4 = m (-4) + 8
⇒ 4 = -4m + 8
and ⇒ 4m = 8 – 4 = 4
⇒ m = 1

Question 9

The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0 ?

Answer  9

P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,
co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 9
Now, substituting x = 0, y = 3 in the equation
x – 5y + 15 = 0
⇒ 0 – 5 x 3 + 15 = 0
and ⇒ 0 – 15 + 15 = 0
⇒ 0 = 0 which is true.
Point (0, 3) lies on the line.

Question 10

The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio of 1 : 2. Does the line x – 2y = 0 contain Q ?

Answer 10

Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2
co-ordinates of Q will be,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A 10
Now, substituting x = 4, y = – 2 in the equation
x – 2y = 0, we get
4 – 2 x (-2) = 0
⇒ 4 + 4 = 0
⇒ 8 = 0 which is not true.
Point Q does not lie on the line x – 2y = 0

Question 11

Find the point of intersection of the lines : 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.

Answer 11

4x + 3y = 1 …..(i)
3x – y = -9 …..(ii).
Multiplying (i) by 1 and (ii) 3
4x + 3y = 1
9x – 3y = -27
Adding, we get-:
1 3x = – 26 ⇒ x = -2
from (ii),
3x – y = – 9
3(-2) – y = -9
⇒ – 6 – y = -9
and ⇒ -y = -9 + 6 = -3
then ⇒ y = 3
The point of intersection is (-2, 3)
The line (2k – 1) x – 2y = 4 passes through that point also
It is satisfy it.
(2k – 1) (-2) – 2(3) = 4
⇒ -4k + 2 – 6 = 4
and ⇒ -4k – 4 = 4
so ⇒ -4k = 4 + 4 = 8
therefore ⇒ k = -2
Hence point of intersection is (-2, 3) and value of k = -2

Question 12

Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.

Answer  12

2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent
They will pass through the same point
Now 2x + 5y = 1 …..(i)
x – 3y = 6 ……(ii)
Multiply (i) by 3 and (ii) by 5, we get :
-6x + 15y = 3
5x – 15y = 30
Adding we get :
11x = 33 ⇒ x = 3
from (ii),
x – 3y = 6
⇒ 3 – 3y = 6
and ⇒ -3y = 6 – 3 = 3
hence ⇒ y = -1
Point of intersection of first two lines is (3, -1)
Substituting the values in third line x + 5y + 2 = 0
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 3 – 5 + 2 = 5 – 5 = 0 = R.H.S.
Hence the given three lines are concurrent.

Concise Solutions of EXERCISE -14 (B), Equation Of A Straight Line Selina Publishers Maths

 

Question 1

Find the slope of the line whose inclination is :
(i) 0°
(ii) 30°
(iii) 72° 30′
(iv) 46°

Answer 1

(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0
(ii) Slope of line whose inclination is 30° = tan 30° = \frac { 1 }{ \surd 3 }
(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)
(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

Question 2

Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875

Answer 2

Slope of a line = tanθ. Where θ is the inclination
(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°
(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°
(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)
(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

Question 3

Find the slope of the line passing through the following pairs of points :
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b ) and (b, -a)

Answer  3

We know that, slope of a line which passes
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 3

Question 4

Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)

Answer 4

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 4

Question 5

Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)

Answer 5

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 5.1

Question 6

The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Answer 6

Slope of the line passing through two points (0, 2) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 6

Question 7

The line passing through (-4, -2) and (2, -3) is perpendicular

Answer 7

Slope of the line passing through the points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 7.1

Question 8

Without using the distance formula, show that the point A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.

Answer 8

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 8
AB and CA are perpendicular to each other
Hence, ΔABC is a right-angled triangle.

Question 9 Equation Of A Line Concise Solutions

Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.

Answer 9

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 9
Slopes of AB and DC are equal
AB || DC Similarly slope of BC and slope of DA are equal.
BC || DA
Hence ABCD is a parallelogram.

Question 10

(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the – quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Answer 10

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 10.3
QR || PS.
Hence PQRS is a parallelogram.

Question 11 Equation Of A Line Concise Solutions

Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are co llinear.

Answer 11

The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)
We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 11
Slope of PQ = Slope of QR
P, Q and R are collinear.

Question 12

Find x, if the slope of the line joining (x, 2) and (8, -11) is \frac { -3 }{ 4 }.

Answer 12

Slope of line joining (x, 2) and (8, -11) is
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 12

Question 13

The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 13

Answer 13 Equation Of A Line Concise Solutions

ΔABC is an equilateral
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 13.1
Each angle is equal to 60°
Side AB is parallel to x-axis
Slope of AB = slope of x-axis = 0.
and Slope of AC = tan A = tan 60° = √3
therefore Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3
hence Slopes of AB, BC and CA are 0, -√3, √3

Question 14

The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :
(i) the slope of the diagonal AC
(ii) the slope of the diagonal BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 14

Answer 14

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 14.1
ABCD is a square in which AB || DC || x-axis.
AD || BC || y-axis
Slope of AB and DC = 0
and slope of AD and BC = not defined (tan90° is not defined)
AC and BD are the diagonals of square ABCD.
Now slope of AC = tan 45° = 1
and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

Question 15

A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :
(i) the slope of the altitude of AB
(ii)  slope of the median AD and
(iii) the slope of the line parallel to AC.

Answer 15

Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 15.1

Question 16

The slope of the side BC of a rectangle ABCD is \frac { 2 }{ 3 }. Find
(i) The slope of the side AB,
(ii) the slope of the side AD.

Answer 16

ABCD is a rectangle in which
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 16

Question 17 Equation Of A Line Concise Solutions

Find the slope and the inclination of the line AB if
(i) A = (-3, -2) and B = (1, 2)
(ii) A = (0, -√3) and B = (3, 0)
(iii) A = (-1, 2√3) and B = (-2, √3)

Answer 17

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 17

Question 18

The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.

Answer 18

Points are collinear.
Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)
Now, Slope of (-3, 2) and (2, -1) will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 18.1

Question 19

The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.
Solution:

Answer 19

Points (k, 3), (2, -4) and (-k + 1, -2) are collinear
Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)
Now, slope of (k, 3) and (2, -4)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 19M

Question 20 Equation Of A Line Concise Solutions

Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC ?
Justify your conclusion by calculating the slopes of AB and AC.

Answer 20

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 20

Question 21

Find the value(s) of k so that PQ will be parallel to RS. Given :

(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)
(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)
(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)

Answer 21

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 21
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 21.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B 21.2

Maths Solutions of Concise for EXERCISE – 14 (C), Equations Of A Straight Line

 

Question 1.

Find the equation of line whose :
(i) y-intercept = 2 and slope = 3,
(ii) y-intercept = -1 and slope = \frac { -3 }{ 4 }

Answer 1

(i) The point whose y-intercept = 2 will be (0, 2) and slope (m) = 3.
Equation of line will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 1.1

Question 2.

Find the equation of a line whose :

(i) y-intercept = -1 and inclination = 45°
(ii) y-intercept = 3 and inclination = 30°

Answer 2

(i) The point whose y-intercept is -1, will be (0, -1) and inclination = 45°
Slope (m) = tan 45° = 1
Equation will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 2

Question 3.

Find the equation of the line whose slope is \frac { -4 }{ 3 } and which passes through (-3, 4).

Answer 3

Slope of the line (m) = \frac { -4 }{ 3 }
The point from which the line passes (-3, 4)
Equation of line will be y – y1 = m (x – x1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 3

Question 4. Equation Of A Line Concise Solutions

Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.

Answer 4

The line passes through the point (5, 4) and angle of inclination = 60°
slope (m) = tan 60° = √3
Equation of line
y – y1 = m (x – x1)
and ⇒ y – 4 = √3 (x – 5)
hence ⇒ y – 4 = √3 x – 5√3
so ⇒ y = √3 x + 4 – 5√3

Question 5.

Find the equation of the line passing through:
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0)
(iii) (4, -2) and (5, 2)

Answer 5

Two given points are (0, 1) and (1, 2)
Slope of the line passing through these two
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 5.1

Question 6. Equation Of A Line Concise Solutions

The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :

(i) The gradient of PQ
and (ii) The equation of PQ,
(iii) The co-ordinates of the point where PQ intersects the x-axis.

Answer 6

Two points P (2,-6) and Q (-3, 5) are given.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 6
and ⇒ 5y – 30 = x – 2
so ⇒ 5y = x – 2 + 30
hence ⇒ 5y = x + 28 ….(i)
(iii) Co-ordinates of the point where PQ intersects x-axis will be = 0
substituting, the value of y in (i)
5 x 0 = x + 28 ⇒ x + 28 = 0 ⇒ x = -28
Co-ordinates of point are (-28, 0)

Question 7.

The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :
(i) the equation of AB
(ii) the co-ordinates of the point where the line AB intersects they- axis.

Answer 7

Slope of the line passing through two points A (-3, 4) and B (2, -1) will be :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 7
Its abscissa = 0
substituting, the value of x = 0 in (i)
0 + y = 1
y = 1
Co-ordinates of point = (0, 1)

Question 8.

The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 8

Answer 8

Two lines AB and CD intersect each other at P (3, 4)
AB inclined at angle of 45° and CD at angle of 60° with the x-axis.
Slope of AB = tan 45° = 1
and slope of CD = tan 60° = √3
Now, equation of line AB will be
y – y1 = m (x – x1)
and ⇒ y – 4 = 1 (x – 3)
⇒ y – 4 = x – 3
so ⇒ y = x – 3 + 4
therefore ⇒ y = x + 1
(ii) Equation of CD will be :
y – y1 = m (x – x1)
and ⇒ y – 4 = √3 (x – 3)
so ⇒ y – 4 = √3 x – 3√3
therefore ⇒ y = √3 x – 3√3 + 4
⇒ y = √3 x + 4 – 3√3

Question 9.Equation Of A Line Concise Solutions

In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.

Answer 9

AD is median
D is mid point of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 9
Equation of AD
y – y1 = m (x – x1)
and ⇒ y + 1 = -6 (x – 4)
so ⇒ y + 1 = -6x + 24
hence ⇒ y + 6x = -1 + 24
⇒ 6x + y = 23

Question 10.

The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex, C = (7, 5). Find the equations of BC and CD.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 10

Answer 10

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 10.1
ABCD is a ||gm in which AB = CD || x-axis
∠A = 60° and C (7, 5)
(i) CD || AB || x-axis ,
Equation of CD will be
y – y1 = m (x – x1)
and ⇒ y – 5 = 0 (x – 7)
so ⇒ y – 5 = 0
hence ⇒ y = 5
BC || AD
Slope of BC = tan 60° = √3
Equation of BC will be
y – y1 = m (x – x1)
and ⇒ y – 6 = √3 (x – 7)
so ⇒ y – 6 = √3 x – 7√3
therefore ⇒ y = √3 x + 6 – 7√3

Question 11.

Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.

Answer 11

Point of intersection of two lines
x + 2y = 7 ….(i)
x – y = 4 ….(ii)
Subtracting, we get
3y = 3
y =1
Substituting, the value of y in (ii)
x – 1 = 4
⇒ x = 4 + 1 = 5
Point of intersection is (5, 1)
Slope of the line passing through origin (0, 0) and (5, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 11
Equation of line will be
y – y1 = m (x – x1)
⇒ y – 5 = \frac { 1 }{ 2 } (x – 1)
and ⇒ 5y – 25 = x – 1
so ⇒ 5y = x – 1 + 25 = x + 24
therefore ⇒ 5y = x + 24

Question 12. Equation Of A Line Concise Solutions

In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.

Answer 12

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 12.1

Question 13.

A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.

Answer 13

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 13
Slope of line through A, perpendicular to BC = -(-1) = 1
Now, the equation of line through A (0, 3) is
y – y1 = m (x – x1)
y – 3 = 1 (x – 0)
and ⇒ y – 3 = x
⇒ y = x + 3

Question 14.

Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).

Answer 14

Slope of the line joining the points (1, 4) and (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 14
Slope of line perpendicular to the above line = 1
Equation of line passing through (-1, 2)
y – y1 = m (x – x1)
⇒ y – 2 = 1 [x -(-1)]
and ⇒ y – 2 = x + 1
so  y = x + 1 + 2
therefore ⇒ y = x + 3

Question 15. Equation Of A Line Concise Solutions

Find the equation of the line, whose :
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
(iv) x-intercept = 3 and y-intercept = -6

Answer 15

(i) When x-intercept = 5, then point will be (5, 0)
and when y-intercept = 3, then point will be (0, 3)
Slope of the line passing through these points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 15.1

Question 16.

Find the equation of the line whose slope is \frac { -5 }{ 6 } and x-intercept is 6.

Answer 16

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 16

Question 17. Equation Of A Line Concise Solutions

Find the equation of the line with x-intercept 5 and a point on it (-3, 2).

Answer 17

x-intercept of the line = 5
Point = (5, 0)
Slope of the line passing through the point (-3, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 17

Question 18.

Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.

Answer 18

The line makes y-intercept = 5
Point = (0, 5)
Slope of the line passing through the point (1, 3) and (0, 5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 18
Equation of the line
y – y1 = m (x – x1)
and ⇒ y – 3 = -2 (x – 1)
so ⇒ y – 3 = -2x + 2
hence ⇒ 2x + y = 2 + 3
⇒ 2x + y = 5

Question 19.Equation Of A Line Concise Solutions

Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 19
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 19.1

Answer 19

(i) Slope of line AB = tan 45° = 1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = 1 (x + 2)
and⇒ y = x + 2
⇒ x – y + 2 = 0
(ii) Slope of line CD = tan (-45°) = -1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = -1 (x + 2)
and ⇒ y = -x – 2
so ⇒ y + x + 2 = 0
hence⇒ x + y + 2 = 0

Question 20. Equation Of A Line Concise Solutions

The line through P (5, 3) intersects y axis at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 20
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.

Answer 20

(i) Here θ = 45°
So, slope of the line = tanθ = tan 45° = 1
(ii) Equation of the line through P and Q is
y – 3 = 1 (x – 5)
⇒ y – x + 2 = 0
(iii) Let the coordinates of Q be (0, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 20.1

Question 21. Equation Of A Line Concise Solutions

Write down the equation of the line whose gradient is \frac { -2 }{ 5 } and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.

Answer 21

Slope of the line m = \frac { -2 }{ 5 }
P divides the line AB, whose co-ordinates are (4, -8) and (12, 0) in the ratio of 3 : 1
Co-ordinates of P be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 21
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 21.1

Question 22.

A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :
(i) the co-ordinates of the centroid of ΔABC.
(ii) the equation of a line, through the centroid and parallel to AB. [2002]

Answer 22

(i) Co-ordinates of vertices of ΔABC are A (1, 4), B (3, 2), C (7, 5)
and let G be the centroid of ΔABC.
Co-ordinates of G are
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 22

Question 23. Equation Of A Line Concise Solutions

A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.

Answer 23

P divides AC in the ratio of 2 : 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C 23

Selina Concise Maths Solutions of Equations of a Straight Line ,EXERCISE – 14(D)

 

Question 1.

Find the slope and y-intercept of the line :
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5

Answer 1

(i) y = 4 ⇒ y = 0x + 4
Here slope = 0 and y-intercept = 4
(ii) ax – by = 0
⇒ by = ax
⇒ y = \frac { a }{ b } x + 0
Here, slope = \frac { a }{ b } and y-intercept = 0
(iii) 3x – 4y = 5
⇒ – 4y = 5 – 3x
and ⇒ 4y = 3x – 5
⇒ y = \frac { 3 }{ 4 } x + \frac { -5 }{ 4 }
Here, slope = \frac { 3 }{ 4 } and y- intercept = \frac { -5 }{ 4 }

Question 2.

The equation of a line is x – y = 4. Find its – slope and y-intercept. Also, find its inclination.

Answer 2

x – y = 4
writing the equation in form of y = mx + c
x = 4 + y
⇒ y = x – 4
Slope = 1 and y-intercept = – 4
Slope = 1
⇒ tanθ = 1
⇒ θ = 45°

Question 3.

(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0 ?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7 ?

and (iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1 ?

(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.

Answer 3 Equation Of A Line Concise Solutions

(i) 3x + 4y + 7 = 0
Writing the equation in form of y = mx + c
4y = -3x – 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 3.2

Question 4.

Find the slope of the line which is parallel to:

(i) x + 2y + 3 = 0
(ii) \frac { x }{ 2 } – \frac { y }{ 3 } – 1 = 0

Answer 4

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 4.1

Question 5.

Find the slope of the line which is perpendicular to:
(i) x – \frac { y }{ 2 } + 3 = 0
(ii) \frac { x }{ 3 } – 2y = 4

Answer 5

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 5.1

Question 6.Equation Of A Line Concise Solutions

(i) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x- ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.

Answer 6

(i) Writing the given equations in the form of y = mx + c, we get:
-by = -2x -5
by = 2x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 6.1

Question 7.

Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

Answer 7

Writing the given equations 2x – y + 5 = 0 and px + 3y = 4 in form of y = mx + c
2x – y + 5 = 0
-y = -2x -5
y = 2x + 5
Here, slope of the line = 2
Again, px + 3y = 4
3y = – px + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 7

Question 8.

The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.

Answer 8

The line AB is given.2* – 2y + 3 = 0
Writing it in the form of y = mx + c
-2y = -2x – 3
and ⇒ 2y = 2x + 3
⇒ y = x + \frac { 3 }{ 2 }
Here, slope of the line = 1
Angle of inclination = tanθ
tanθ = 1
θ = 45°

Question 9.Equation Of A Line Concise Solutions

The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.

Answer 9

Writing the given lines 4x + 3y = 9 and px – 6y + 3 = 0 in the form of y = mx + c
4x + 3y = 9
⇒ 3y = – 4x + 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 9

Question 10.

If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)

Answer 10

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 10.1

Question 11.

The line through A (-2, 3) and B (4, 6) is perpendicular to the line 2x – 4y = 5. Find the value of b.

Answer 11

Gradient (mx) of the line passing through the points A (-2, 3) and B (4, b)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 11

Question 12.

Find the equation of the line passing through (-5, 7) and parallel to
(i) x-axis
(ii) y-axis.

Answer 12

(i) Slope of the line parallel to x-axis = 0
Equation of line passing through (-5, 7) whose slope is 0.
y – 7 = 0 [x – (-5)]
and ⇒ y – 7 = 0
⇒ y = 7
(ii) Slope of the line parallel to y-axis = 0
y – y1 = m (x – x1)
and ⇒ 0 = x – x1
⇒ x + 5 = 0

Question 13. Equation Of A Line Concise Solutions

(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1). (2007)

Answer 13

(i) Writing the equation x – 3y = 4 in form of y = mx + c
-3y = -x + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 13
⇒ 2y – 2 = -3x
⇒ 3x + 2y – 2 = 0
Which is the required equation.

Question 14.

Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Answer 14

Writing the equation 4x + 5y = 6 in form of y = mx + c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 14
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 14.1

Question 15.

Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).

Answer 15 Equation Of A Line Concise Solutions

The perpendicular of the line segment bisects it.
Co-ordinates of mid-point of the line segment which is obtained by joining the points (6, -3) and (0, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 15
Slope of the line perpendicular to it = 1 (Product of slopes = -1)
Equation of the perpendicular bisector is y – y1 = m (x – x1)
and y – 0 = 1 (x – 3)
y = x – 3

Question 16.

In the following diagram, write down:
(i) the co-ordinates of the points A, Band C.
(ii) the equation of the line through A and parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 16

Answer 16

(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 16.1

Question 17.

B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.

Answer 17

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 17

Question 18.

A = (7, -2) and C = (-1, – 6) are the vertices of a square ABCD. Find the equation of the diagonals AC and BD.

Answer 18

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 18.1

Question 19. Equation Of A Line Concise Solutions

A (1, -5), B (2, 2) and C (-2, 4) are the vertices of the ∆ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.

Answer 19

(i) Let D be the mid-point of BC
co-ordinates mid-point of
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 19
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 19.1

Question 20.

(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin. [1995]

Answer 20

(i) Write the equation 2y = 3x + 5 in the form of y = mx + c. We get:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 20
y – 2 = \frac { -2 }{ 3 } (x – 3)
⇒ 3y – 6 = -2x + 6
and ⇒ 2x + 3y = 6 + 6
⇒ 2x + 3y = 12 …… (i)

(ii)

AB meets the x-axis at A
ordinate (y) of A = 0 i.e. y = 0
Substituting, the value of y in (i)
2x + 3 x 0 = 12
⇒ 2x = 12
⇒ x = 6
Co-ordinates of A are (6, 0)
Again. AB meets y-axis at B
Abscissa of B = 0 i.e. x = 0
Substituting the value of x in (i)
2 x 0 + 3y = 12
⇒ y = 4
Co-ordinates of B are (0, 4)
Area of ∆OAB = \frac { 1 }{ 2 } x Base x altitude
\frac { 1 }{ 2 } x 4 x 6 = 12 square units

Question 21. Equation Of A Line Concise Solutions

The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.

Answer 21

The line 4x – 3y + 12 = 0 meets x-axis at A.
Ordinates of A = 0. i.e. y = 0
Substituting, the value of y in the equation
4x – 3 x 0 + 12 = 0
⇒ 4x + 12 = 0
and ⇒ 4x = -12
hence ⇒ x = -3
Co-ordinates of A are (-3, 0)
Writing the equation 4x – 3y + 12 = 0 in form of y = mx + c
⇒ -3y = -4x – 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 21
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 21.1

Question 22.

The point P is the foot of perpendicular from A (-5, 7) to the line is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP.
(ii) the co-ordinates of P.

Answer 22

Write the equation in form of y = mx + c
2x – 3y + 18 = 0
⇒ -3y = -2x – 18
⇒ y = \frac { 2 }{ 3 } x + 6 (Dividing by 3)
Slope of the line = \frac { 2 }{ 3 }
and slope of the line perpendicular to it = \frac { -3 }{ 2 } (Product of slopes = -1)

(i)

Equation of line AP perpendicular to the given line and drawn through A (-5, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 22
y – y1 = m (x – x1)
⇒ y – 7 = \frac { -3 }{ 2 } (x + 5)
⇒ 2y – 14 = -3x – 15
and ⇒ 3x + 2y – 14 + 15 = 0
hence ⇒ 3x + 2y + 1 = 0

(ii)

P is the point of intersection of these lines
we will solve their equations
2x – 3y + 18 = 0 ….(i)
3x + 2y + 1 = 0 ….(ii)
Multiplying (i) by 2 and (ii) by 3, we get
4x – 6y + 36 = 0
9x + 6y + 3 = 0
Adding, we get:
13x + 39 = 0
and ⇒ 13x = -39
⇒ x = -3
Now, substituting the value of x in (i)
2(-3) – 3y + 18 = 0
⇒ -6 – 3y + 18 = 0
and ⇒ -3y + 18 – 6 = 0
so ⇒ -3y + 12 = 0
hence⇒ -3y = -12
⇒ 3y = 12
⇒ y = 4
Co-ordinates of P are (-3, 4)


Question 23

The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC. If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.

Answer 23

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 23
AB meets y- axis at P
abscissa of P = 0 i. e. x = 0
Substituting the value of y in (i)
0 + y = 4
⇒ y = 4
Co-ordinates of P are (0, 4)
BC meets x-axis at Q
ordinate of Q = 0 i.e. y = 0
Substituting, the value of y in (ii),
2x + 0 = 6
⇒ 2x = 6
⇒ x = 3
Co-ordinates of Q are (3, 0)

 

Question 24

Match the equations A, B, C, and D with the lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram.
A = y = 2x;
B = y – 2x + 2 = 0;
C = 3x + 2y = 6;
D = y = 2 [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 24

Answer 24

A → L3,
B → L4,
C → L2,
D → L1

Question 25.

Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. (2014)

Answer 25

A (a, 3), B (2,1) and C (5, a) are collinear.
Slope of AB = Slope of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D 25

EXERCISE – 14 (E) Solutions of Concise Selina Publications Maths Equation of a Straight Line

 

Question 1.

Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7

Answer 1

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 1.1

Question 2

The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.

Answer 2

Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3
Let co-ordinates of P be (x, y), then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 2.1

Question 3

Question 3.

A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.

Answer 3

P lies on y-axis and let the co-ordinates of P be (0, y)
P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.
5 x 0 + 3y + 15 = 0
⇒ 3y = -15
⇒ y = -5
Co-ordinates of P are (0, -5)
Now, writing the line x – 3y + 4 = 0 is form of y = mx + c
-3y = -x – 4
and ⇒ 3y = x + 4
⇒ y = \frac { 1 }{ 3 } x + \frac { 4 }{ 3 }
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 3

Question 4

Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]

Answer 4

Writing, the line kx – 5y + 4 = 0 in form of y = mx + c
⇒ -5y = -kx – 4
⇒ 5y = kx + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 4

Question 5

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M. (2003)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 5

Answer 5

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 5.1
(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.
Substituting, the co-ordinates in (i)
x + 0 = 3 ⇒x = 3
Co-ordinates of A are (3, 0)
Again 0 + y = 3 ⇒ y = 3
Co-ordinates of B are (0,3)
(iii) M is the mid-point of AB.
Co-ordinates of M wil be (\frac { 3 }{ 2 } , \frac { 3 }{ 2 })

Question 6

(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.

Answer 6

Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 6.1
⇒ 3y – 6 = -2x – 2
and ⇒ 2x + 3y = 6 – 2
so ⇒ 2x + 3y = 4

Question 7

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Answer 7

Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)
Let, co-ordinates of fourth vertex D be (x, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 7.3

Question 8

.A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.

Answer 8

Slope of line x = 3y + 2 or 3y = x – 2 ….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 8.1

Question 9

A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer 9

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 9
Let, the line intersects x-axis at A and y-axis at B.
Let, co-ordinates of A (x, o) and of (o, y)
But (3, 2) is the mid-point of AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 9.1

Question 10

Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.

Answer 10

7x + 6y = 71 ….(i)
5x – 8y = -23 ….(ii)
Multiply (i) by 4 and (ii) by 3,
28x + 24y = 284
15x – 24y = -69
On adding (i) and (ii), we get:
43x = 215
x = 5
Substituting, the value of x in (i)
7 x 5 + 6y = 71
35 + 6y = 71
⇒ 6y = 71 – 35 = 36
⇒ y = 6
Point of intersection of these lines is (5, 6)
Now slope of line 4x – 2y = 1
⇒ 4x – 1 = 2y
⇒ y = 2x – \frac { 1 }{ 2 } is 2
Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is \frac { -1 }{ 2 }
Equation of the line y – y1 = m (x – x1)
⇒ y – 6 = \frac { -1 }{ 2 } (x – 5)
and ⇒ 2y – 12 = -x + 5
hence ⇒ x + 2y = 5 + 12 = 17
⇒ x + 2y = 17

Question 11

Find the equation of the line which is perpendicular to the line \frac { x }{ a } – \frac { y }{ b } = 1 at the point where this line meets y-axis.

Answer 11

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 11
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 11.1

Question 12

O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of ∆OAB through vertex O.
(ii) the equation of altitude of ∆OAB through vertex B.

Answer 12

(i) Let, mid-point of AB be D.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 12.2

Question 13

Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does line 3x = y + 1 bisect the line segment joining the two given points ?

Answer 13

Slope of the line joining the points (-2, 3) and (4, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 13
Yes, these are perpendicular to each other
Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 13.1
This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point
now, substituting the value of x and y. in 3x = y + 1
⇒ 3 x 1 = 2 + 1
⇒ 3 = 3. which is true.
Yes, the line 3x = y + 1 is the bisector.

Question 14

Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).

Answer 14

Equation of the given line is x cos 30° + y sin 30° = 2
y sin 36° = -x cos 30° + 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 14

Question 15

Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.

Answer 15

Writing the given equation in the form of y = mx + c
(k – 2) x + (k + 3) y – 5 = 0 ….(i)
⇒ (k + 3) y = – (k – 2) x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 15.1

Question 16

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.

Answer 16

Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 16
(ii) Let, the line through A meets BC in P
P is point of intersection of these two lines.
3x – 4y = 5 ……… (i)
4x + 3y = 15 …….. (ii)
On solving (i), (ii) we get
x = 3, y = 1
Co-ordinates of Pare (3, 1)

Question 17

From the given figure, find :
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC. (2005)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 17

Answer 17

(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)
(ii) Slope of line BC is (m)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 17.1
⇒ x + 2y – 6 – 2 = 0
and ⇒ x + 2y – 8 = 0
hence ⇒ x + 2y = 8

Question 18

P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)

Answer 18

Let (x, y) be the co-ordinates of M, the mid-point of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 18.1
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 19

A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.

Answer 19

In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.
P and Q are the mid-points of AB and AC respectively
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 19
Slopes of PQ and BC are same.
These are parallel to each other.
Quad. PBCQ is trapezium

Question 20

.A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :
(i) the co-ordinates of A and B.
(ii) equation of line through P, and perpendicular to AB.

Answer 20

Line AB intersects x-axis at A and y-axis at B.
(i) Let co-ordinates of A be (x, 0) and of B be (0, y)
Point P (-4, -2) intersects AB in the ratio 1 : 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 20
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 20.1

Question 21.

A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)

Answer 21

Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 21

Question 22.

Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)

Answer 22

x-intercept = 4
Co-ordinates of that point = (4, 0)
The co-ordinates of the given point (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 22

Question 23.

The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 23

Answer 23

Equation of line AB is y = x + 1
and equation of line CD is y = √3 x – 1
Slope of AB = 1
tanθ = 1
⇒ θ = 45°
Inclination angles of AB = 45°
Slope of CD = tanθ = √3 = tan 60°
⇒ θ = 60°
Inclination angle of CD = 60°
In ΔPQR,
Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)
⇒ 60° = 45° + θ
and ⇒ θ = 60° – 45° = 15°
hence ⇒ θ = 15°

Question 24.

Write down the equation of the line whose gradient is \frac { 3 }{ 2 } and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)

Answer 24

P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 24

Question 25.

The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.

Answer 25

Let points are A (6, 4) and B (7, -5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 25

Question 26.

Points A and B have coordinates (7, -3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ of (-2, p) lies on it.

Answer 26

Coordinates of A are (7, -3), of B = ( 1, 9)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 26

Question 27.

A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid point of AB. Find the
(i) Coordinates of A and B.
(ii) Slope of line AB.
(iii) equation of line AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 27

Answer 27

As P (2, -3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 27.1

Question 28.

The equation of a line is 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.

Answer 28

Given line 3x + 4y -7 = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 28
⇒ 3 (y – 4) = 4 (x – 2)
and ⇒ 3y – 12 = 4x – 8
so ⇒ 4x – 3y – 8 + 12 = 0
hence ⇒ 4x – 3y + 4 = 0

Question 29.

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :
(i) co-ordinates of A
(ii) equation of diagonal BD.

Answer 29

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 29
(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)
The diagonals of ||gm bisect each other
O is said point of AC and BD
Now if O is mid point of BD then its co-ordinates will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 29.1
⇒ y + 4 = 4x – 8
and ⇒ 4x – y -8 – 4 = 0
hence ⇒ 4x – y – 12 = 0 or 4x – y = 12

Question 30.

Given equation of line L1 is y = 4.
(i) Write the slope of line L2 if L2 is the bisector of angle O.
(ii) Write the co-ordinates of point P.
(iii) Find the equation of L2.

Answer 30

(i) Equation of line L1 is y = 4
L2 is the bisector of ∠O
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 30
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 30.1

Question 31.

Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 31
(i) equation of AB
(ii) equation of CD

Answer 31

Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 31.1

Question 32.

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1

Answer 32

x-intercept of the line = -3
and is perpendicular to the line
3x + 5y = 1
5y = 1 – 3x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 32

Question 33.

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M.

Answer 33

A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.
M is mid-point of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 33
(i) Equation of line AB will be
y – y1 = m (x – x1)
⇒ y – 4 = -1 (x + 1)
and ⇒ y – 4 = – x – 1
therefore⇒ y + x = -1 + 4
hence ⇒ x + y = 3
(ii) The line intersect x-axis at A and OA = 3 units
Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units
Co-ordinates of B are (0, 3)
(iii) M is mid-point of AB
Co-ordinates of M are (\frac { 3 }{ 2 } , \frac { 3 }{ 2 })

Question 34.

In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 34
(i) equation of line AB.
(ii) equation of line CD.
(iii) co-ordinates of points E and D.

Answer 34

In the given figure, AB meets y-axis at point A.
Line through C (2, 10) and D intersects line AB at P at right angle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 34.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 34.2
Equation of CD
y – 10 = 3 (x – 2)
⇒ y – 10 = 3x – 6
and ⇒ 3x – y + 10 – 6 = 0
so ⇒ 3x – y + 4 = 0
(iii) Co-ordinates of D which is on x-axis
3x – y + 4 = 0
3x – 0 + 4 = 0
⇒ 3x + 4 = 0
and ⇒ 3x = -4
hence ⇒ x = \frac { -4 }{ 3 }
Co-ordinates of D are (\frac { -4 }{ 3 } , 0)
E is also on x-axis
x + 3y = 18
Substituting, y = 0, then
x + 0= 18
⇒ x = 18
Co-ordinates of E are (18, 6)

Question 35.

A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.

Answer 35

A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.
Draw BC || x-axis
and PC || y-axis
In ∆PAD and ∆PBC
∠P = ∠P (common)
∠D = ∠C (each 90°)
∆PAD ~ ∆PBC
PB = 2PA ⇒ PA = \frac { 1 }{ 2 } PB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 35
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 35.1
2y – 6 = 3x – 12
⇒ 3x – 2y – 12 + 6 = 0
and ⇒ 3x – 2y – 6 = 0
hence ⇒ 3x – 2y = 6

Question 36.

Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.

Answer 36

Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 36
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 36.1

Question 37.

Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.

Answer 37

The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)
Now slope of the line joining A and B
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 37
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 37.1
Equation of the line y – y1 = m(x – x1)
⇒ y – 2 = -1 (x – 0)
and ⇒ y – 2 = -x
hence ⇒ x + y – 2 = 0
Point C (1, 1) will be on AB if it satisfy
1 + 1 – 2 = 0
⇒ 0 = 0
Point C lies on AB
Hence points A, C and B are collinear.

Question 38.

Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)

Answer 38

Three vertices of a ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)
Join diagonals AC and BD which bisect each other at O.
O is mid-point of AC as well as of BD
Now co-ordinates of O will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 38
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 38.1

Question 39.

In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 39
(i) Write the co-ordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC. (2015)

Answer 39

In the given figure,
ABC is a triangle and BC || y-axis
AB and AC intersect the y-axis at P and Q respectively.
(i) Co-ordinates of A are (4,0).
(ii) Length of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 39.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 39.2

Question 40.

The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \frac { 1 }{ 2 }. Find :
(i) k
(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)

Answer40

(i) Slope of the line joining P(6, k) and Q (1 –
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 40

Question 41.

A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 41

Answer 41

(i) Since, A lies on the x-axis,
let the coordinates of A be (x, 0).
Since B lies on the y-axis,
let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 41.1

End of Chapter-14 Equation of a Line Concise Maths for Class 10th

 

Chapter Wise Concise Maths Solution for ICSE Class 10th

  1. GST ( Goods and Services Tax)
  2. Banking
  3. Shares and Dividends
  4. Linear In equations in One Variable
  5. Quadratic Equations 
  6. Solving Simple Problems (Based on Quadratic Equations)
  7. Ratio and Proportion ( Including Properties and Uses )
  8. Reminder and Factor Theorems
  9. Matrices
  10. Arithmetic Progression
  11. Geometric Progression
  12. Reflection
  13. Section And Mid Point Formula 
  14. Equation of a Line (Currently Open)
  15. Similarity (With Applications to Maps and Models )
  16. Loci (Locus and its Constructions)
  17. Circles
  18. Tangents and Intersecting Chords
  19. Constructions ( Circles)
  20. Cylinder, Cone and Spheres (Surface Area and Volume )
  21. Trigonometrical Identities
  22. Heights and Distances
  23. Graphical Representation (Histograms and Ogives)
  24. Measures of Central Tendency (Mean , Median, Quartiles and Mode)
  25. Probability

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